# Fundamental Group of Connected Sum of 3 Tori by Van Kampen Theorem I cut the surface in two open subsets; one of them, say, $$A$$, a $$2$$-torus with a hole on its side and the other, say $$B$$, a torus with a hole on its side. The first deformation retracts to $$4$$ circles attached end-to-end (or the wedge sum of two $$2$$-circles) and the other deformation retracts to $$2$$ circles attached end-to-end. Their intersection deformation retracts to a single circle. Let $$\pi_1(A)=\langle a,b,c,d\rangle$$, $$\pi_1(B)=\langle e,f\rangle$$ and $$\pi_1(A\cap B)=\langle w\rangle$$. Now $$i_{B}(w)=efe^{-1}f^{-1}$$. But I cannot figure out what $$i_{A}(w)$$ should be. I guess it is something like $$aba^{-1}b^{-1}cdc^{-1}d^{-1}$$ but I don't know why.

Consider a 1-genus torus with two boundary components: and the paths $$a,b,w_1,w_2$$ as shown here: Since you say

Now $$i_B(w) = efe^{-1}f^{-1}$$

I'll assume you are somewhat comfortable with the fact that the boundary of $$B$$ (the torus with a single hole) can be expressed as $$efe^{-1}f^{-1}$$. It is a similar fact that $$w_1$$ and $$w_2$$ differ by the commutator $$[a,b] = aba^{-1}b^{-1}$$. That is, $$w_1aba^{-1}b^{-1} = w_2$$. I encourage you to draw this all out and convince yourself of this. (I may have drawn the orientations backwards, so you may have to throw in some inverses to make it work.)

Now consider $$A$$: Here, I've drawn in some additional curves to help us out. Strictly speaking, these should all be attached to some basepoint as they were in the previous picture, but I didn't want to clutter it. Also for the sake of clutter reduction, I've omitted $$a,b,c,d$$, where $$a,b$$ are the paths as in the previous picture around the leftmost hole, and $$c,d$$ are those around the rightmost.

By applying the first paragraph to the leftmost hole, we know that $$w_1aba^{-1}b^{-1} = w_2$$. Similarly, for the rightmost hole, we have $$w_1 = wcdc^{-1}d{-1}$$. Substituting, we get $$w_2 = wcdc^{-1}d{-1}aba^{-1}b^{-1}$$. But $$w_2$$ is homotopically trivial, so $$w = bab^{-1}a^{-1}dcd^{-1}c^{-1}$$.

As before, this depends on exactly what the orientations of everything are, so you might have to throw in some inverses. Accounting for that, (I think) this is what you wanted.

Hope it helps.

I'll also mention that the usual (and easier) way of finding the fundamental group of the $$n$$-genus torus is by using the $$4n$$-gon identification, an outline of which can be found here.